Algebra Pre-Algebra and Basic Algebra Math Forum

 April 28th, 2015, 12:25 PM #1 Newbie   Joined: Apr 2015 From: England Posts: 6 Thanks: 0 Please help me to understand I have recently stumbled across this following question: The answers are the following: I would like to know how to work out the answers as I got them wrong. i) I understand why it would be xy, but I don't understand why if it is xy the number would be 2^p+s and not 4^p+s. ii) I completely don't understand this. iii) Completely confused with this one as well. It would be great if someone could help and say how those answers came about. Thanks a lot. Last edited by skipjack; April 28th, 2015 at 02:57 PM.
 April 28th, 2015, 12:33 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond i) $\displaystyle 2^{p+s}=(2^p)(2^s)=xy$ ii) $\displaystyle 2^{2p}=(2^p)^2=x^2$ iii) $\displaystyle 2^{p-1}=(2^p)(2^{-1})=2^p\cdot\frac12=x\cdot\frac12=\frac12x$ Thanks from aurel5 and nautudent
 April 28th, 2015, 12:50 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,446 Thanks: 2499 Math Focus: Mainly analysis and algebra Recall that (for any positive integers $p$ and $s$) $$2^p = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p terms}} \\ 2^s = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{s terms}}$$ So when we multiply we get $$2^p \times 2^s = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p terms}} \times \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{s terms}} = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p+s terms}} = 2^{p+s}$$ Similarly, $$\left(2^p\right)^2 = 2^p \times 2^p = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p terms}} \times \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p terms}} = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p+p terms}} = 2^{p+p} = 2^{2p}$$ Finally, $$2^{p-1} \times 2 = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p-1 terms}} \times 2 = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p terms}} = 2^p$$ so we can divide by 2 to get $$2^{p-1} = \frac12 2^p$$ These results generalise to any $p$ and $s$. Thanks from nautudent
April 28th, 2015, 01:03 PM   #4
Newbie

Joined: Apr 2015
From: England

Posts: 6
Thanks: 0

Quote:
 Originally Posted by greg1313 i) $\displaystyle 2^{p+s}=(2^p)(2^s)=xy$ ii) $\displaystyle 2^{2p}=(2^p)^2=x^2$ iii) $\displaystyle 2^{p-1}=(2^p)(2^{-1})=2^p\cdot\frac12=x\cdot\frac12=\frac12x$
Cheers for the working out, very helpful

Quote:
 Originally Posted by v8archie Recall that (for any positive integers $p$ and $s$) $$2^p = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p terms}} \\ 2^s = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{s terms}}$$ So when we multiply we get $$2^p \times 2^s = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p terms}} \times \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{s terms}} = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p+s terms}} = 2^{p+s}$$ Similarly, $$\left(2^p\right)^2 = 2^p \times 2^p = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p terms}} \times \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p terms}} = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p+p terms}} = 2^{p+p} = 2^{2p}$$ Finally, $$2^{p-1} \times 2 = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p-1 terms}} \times 2 = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{p terms}} = 2^p$$ so we can divide by 2 to get $$2^{p-1} = \frac12 2^p$$ These results generalise to any $p$ and $s$.
I understand now, very well explained
thank you very much (:
One more thing, as you said you explained the p+s terms come together to form 2^p+s
But how come when you do a multiplication such as 9^3 x 9^3 = It equals 81^6 and not 9^6

Last edited by nautudent; April 28th, 2015 at 01:07 PM.

April 28th, 2015, 01:26 PM   #5
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,446
Thanks: 2499

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by nautudent But how come when you do a multiplication such as 9^3 x 9^3 = It equals 81^6 and not 9^6
$$9^3 \times 9^3= (9 \times 9 \times 9) \times ( 9 \times 9 \times 9) = \underbrace{9 \times 9 \times 9 \times 9 \times 9 \times 9}_{\text{6 terms}} = 9^6$$

May 2nd, 2015, 04:08 AM   #6
Newbie

Joined: Apr 2015
From: England

Posts: 6
Thanks: 0

Quote:
 Originally Posted by v8archie $$9^3 \times 9^3= (9 \times 9 \times 9) \times ( 9 \times 9 \times 9) = \underbrace{9 \times 9 \times 9 \times 9 \times 9 \times 9}_{\text{6 terms}} = 9^6$$
Hmm so it is 9^6
so why would this be right??

 May 2nd, 2015, 04:34 AM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,446 Thanks: 2499 Math Focus: Mainly analysis and algebra \begin{aligned}4pq \times 6p^2q &= 4 \times p \times q \times 6 \times p \times p \times q \\ &= \underbrace{4 \times 6}_{24} \times \underbrace{p \times p \times p}_{p^3} \times \underbrace{q \times q}_{q^2} \\ &= 24p^3q^2\end{aligned} Thanks from nautudent
 May 2nd, 2015, 04:44 AM #8 Global Moderator   Joined: Dec 2006 Posts: 19,722 Thanks: 1807 One can add the exponents when finding the product of powers of the same thing: $p^mp^n = p^{m+n}$. If the exponents are equal, one can find the product as follows: $p^nq^n = (pq)^n$. Hence $9^39^3$ is equal to both $9^6$ and $81^3$. Thanks from nautudent

 Tags understand

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post bradhills Algebra 1 February 21st, 2013 12:48 PM ment0smintz Calculus 6 February 18th, 2013 12:07 PM goodjobbro Number Theory 13 December 25th, 2012 06:36 PM scrum Calculus 2 January 28th, 2010 12:33 AM johnny Algebra 1 July 21st, 2009 01:18 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top